/* 1118:Lining Up */
/*
描述
"How am I ever going to solve this problem?" said the pilot.

Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?


Your program has to be efficient!
输入
Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.
输出
output one integer for each input case ,representing the largest number of points that all lie on one line.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

const int maxn=1001;
const int inf=1<<28;
struct node
{
    int x;
    int y;
    bool operator < (const node &a)const
    {
    if(x==a.x)
        return y<a.y;
    return x<a.x;
    }
}a[maxn];
int n,ans;
double pos[maxn];
int main()
{
    while(scanf("%d",&n)&&n)
    {
    ans=1;
    for(int i=0;i<n;i++)
        scanf("%d%d",&a[i].x,&a[i].y);
    sort(a,a+n);
    for(int i=0;i<n;i++)
    {
        int cur=0;
        for(int j=i+1;j<n;j++)
        {
        if(a[j].x-a[i].x==0)
            pos[cur++]=inf;
        else
            pos[cur++]=(a[j].y-a[i].y)*1.0/(a[j].x-a[i].x);
        }
        sort(pos,pos+cur);
        int cnt=1;
        for(int j=1;j<cur;j++)
        {
        if(fabs(pos[j]-pos[j-1])<1e-6)
        {
            cnt++;
            ans=max(ans,cnt);
        }
        else
            cnt=1;
        }
    }
    printf("%d\n",ans+1);
    }
    return 0;
}

//	Accepted
